Question

Find the value of ${\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right) =$
( 1) ${\tanh ^{ - 1}}\sqrt 5$
( 2) ${\tanh ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)$
( 3) ${\tanh ^{ - 1}}\sqrt 3$
( 4) ${\tanh ^{ - 1}}\left( {\dfrac{2}{{\sqrt 5 }}} \right)$

Answer 1

First we have to keep in mind that it is inverse $\sin$hyperbolic or ${\sinh ^{ - 1}}$ function ; not inverse $\sin$function . Then we should obtain the value of ${\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ using formula . After finding that value we have to equate that with the value of ${\tanh ^{ - 1}}x$ . We have to use the formula of ${\tanh ^{ - 1}}x$ and find the value of $x$. For simplification and calculation we have to use some basic logarithmic and fraction formulas .
FORMULA USED
${\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$
${\tanh ^{ - 1}}x = \dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$

Complete answer: First we have to evaluate the value of ${\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ by formula .
Putting the value of $x$equal to $\dfrac{1}{2}$we get
${\sinh ^{ - 1}}\dfrac{1}{2} = \ln \left( {\dfrac{1}{2} + \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + 1} } \right)$
$= \ln \left( {\dfrac{1}{2} + \sqrt {\left( {\dfrac{1}{4}} \right) + 1} } \right)$
$= \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)$
So we get ${\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ $= \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)$
Now we have to equate it with the value of ${\tanh ^{ - 1}}x$and after that with the help of a formula we will get the value of $x$.
$\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)$
Now for the simplicity of calculation we will use some logarithmic formula and we can get
$\dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{\dfrac{1}{2}}}$
Then putting these value in previous equation we get
$\ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{\dfrac{1}{2}}} = \ln \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)$
$\Rightarrow {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{\dfrac{1}{2}}} = \left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)$
Taking square on both side we get
$\Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = {\left( {\dfrac{1}{2} + \sqrt {\dfrac{5}{4}} } \right)^2}$
$\Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \left( {{{\left( {\dfrac{1}{2}} \right)}^2} + \dfrac{5}{4} + 2 \times \dfrac{1}{2} \times \dfrac{{\sqrt 5 }}{2}} \right)$
After simplifying we get
$\Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \left( {\dfrac{3}{2} + \dfrac{{\sqrt 5 }}{2}} \right)$
$\Rightarrow \left( {\dfrac{{1 + x}}{{1 - x}}} \right) = \left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)$
After cross multiplication we get
$\Rightarrow 2 \times \left( {1 + x} \right) = \left( {3 + \sqrt 5 } \right) \times \left( {1 - x} \right)$
$\Rightarrow 2 + 2x = 3 - 3x + \sqrt 5 - \sqrt 5 x$
After simplification and taking all $x$ containing term in one side we get
$\Rightarrow 5x + \sqrt 5 x = 1 + \sqrt 5$
$\Rightarrow x \times \left( {5 + \sqrt 5 } \right) = \left( {1 + \sqrt 5 } \right)$
After simplifying we get
$\Rightarrow x = \dfrac{{1 + \sqrt 5 }}{{5 + \sqrt 5 }}$
To rationalise we multiply both numerator and denominator with the conjugate of denominator which is $\left( {5 - \sqrt 5 } \right)$
$\Rightarrow x = \dfrac{{\left( {1 + \sqrt 5 } \right)\left( {5 - \sqrt 5 } \right)}}{{\left( {5 + \sqrt 5 } \right)\left( {5 - \sqrt 5 } \right)}}$
After multiplication we get
$\Rightarrow x = \dfrac{{5 - \sqrt 5 + 5\sqrt 5 - 5}}{{25 - 5}}$
After simplification we get
$\Rightarrow x = \dfrac{{4\sqrt 5 }}{{20}}$
$\Rightarrow x = \dfrac{1}{{\sqrt 5 }}$
So we get the final answer ${\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)$=${\tanh ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right)$
So option 2 is the correct answer .

Note:
Remember we are dealing with hyperbolic function not simple trigonometric function .
We have to keep in mind basic logarithmic and fraction formulas and students can simplify the equation in a way s/he can find easiest . For cross check the answer student can put the value $x = \dfrac{1}{{\sqrt 5 }}$in inverse hyperbolic formula given by ${\tanh ^{ - 1}}x = \dfrac{1}{2}\ln \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$.Alternative but tedious method is to check the value of all four option and see which one is equal to ${\sinh ^{ - 1}}\left( {\dfrac{1}{2}} \right)$.